Solve for x and y:

ax+by=1

bx+ay=(2ab)/(a²+b²)

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Question

Solve for x and y:
ax+by=1
bx+ay=(2ab)/(a2+b2)

Ankush Jain · Accepted Answer

$a x + b y = 1 and b x + a y = \frac{2 a b}{a^{2} + b^{2}}$

The given equations can be rewritten as :

$a x + b y - 1 = 0 and b x + a y - \frac{2 a b}{a^{2} + b^{2}} = 0$

Using method of cross - multiplication, we get

$\frac{x}{\frac{- 2 a b^{2}}{a^{2} + b^{2}} + a} = \frac{- y}{\frac{- 2 a^{2} b}{a^{2} + b^{2}} + b} = \frac{1}{a^{2} - b^{2}} \Rightarrow \frac{x}{\frac{- 2 a b^{2} + a^{3} + a b^{2}}{a^{2} + b^{2}}} = \frac{- y}{\frac{- 2 a^{2} b + a^{2} b + b^{3}}{a^{2} + b^{2}}} = \frac{1}{a^{2} - b^{2}} \Rightarrow \frac{(a^{2} + b^{2}) x}{a^{3} - a b^{2}} = \frac{- y (a^{2} + b^{2})}{b^{3} - a^{2} b} = \frac{1}{a^{2} - b^{2}} \Rightarrow \frac{(a^{2} + b^{2}) x}{a (a^{2} - b^{2})} = \frac{y (a^{2} + b^{2})}{b (a^{2} - b^{2})} = \frac{1}{a^{2} - b^{2}} \Rightarrow \frac{(a^{2} + b^{2}) x}{a (a^{2} - b^{2})} = \frac{1}{a^{2} - b^{2}} and \frac{y (a^{2} + b^{2})}{b (a^{2} - b^{2})} = \frac{1}{a^{2} - b^{2}} \Rightarrow x = \frac{a}{a^{2} + b^{2}} and y = \frac{b}{a^{2} + b^{2}}$

Solve for x and y: ax+by=1 bx+ay=(2ab)/(a2+b2)

Solve for x and y:

ax+by=1

bx+ay=(2ab)/(a²+b²)